See complete Problem 96
Similar Triangles, Incenters, Parallelogram. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Geometry Problem 96
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See complete Problem 96
Similar Triangles, Incenters, Parallelogram. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
join B to O1 & O extend to P ( P on AC )(same bisector)
ReplyDeleteextend GO3 to Q ( Q on AC )
ang APB = 1/2 B + C
ang MQG = 1/2 G + C ( AB//MF => ang B = ang G )
=>
ang APB = ang MQG
=>
PB // QO2
ang DEA = ang C ( DE // AC )
=>1/2 DEA = 1/2 C
=>
O2O1 // CO
To Antonio:
ReplyDeleteThe written enunciate of problem 96 is a repetition of problem 94. The figure is all right.
Thanks Nilton. The written enunciate of problem 96 has been updated.
DeleteAlternative solution to problem 96.
ReplyDeletePoints O, O1 and B are collinear, on the bisector of ang(ABC). Points O2, G and O3 are collinear, on the bisector of ang(MGC). As triangles ABC, DBE, FGE and MGC are similar,
then OB/BC = O1B/BE = O2G/EG = O3G/GC = (O1B+O2G+O3G)/(BE+EG+GC).
But BE + EG + GC = BC, so OB = O1B + O2G + O3G, which is the same as OB – O1B = O2G + O3G. Besides OO1 = OB – O1B = O2G + O3G = O2O3.
Being ABC and BGC similar, then
ang(ABO) = ang(MGO3), and, as AB // MR, then
OO1 // O2O3.
We can see now that OO1O2O3 is a parallelogram because it has two opposite sides equal and parallel.
Same as Problem 93
ReplyDelete